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FP1 dividing complex numbers

for something like

35i2i\dfrac{3-5i}{2i}

why not multiply by 2i/2i why multiply by -2i/-2i instead?
(edited 7 years ago)
Original post by timebent
for something like

35i2i\dfrac{3-5i}{2i}

why not multiply by 2i why multiply by -2i instead?


While the conjugate of 2i is -2i, it is not necessary. As you've suggested you can multiply by 2i2i=1\dfrac{2i}{2i}=1

But where the complex number in the denominator has both a real an imaginary part, you multiply by the conjugate which is abia-bi for a complex number a+bia+bi) as you make use of the difference of two squares (a2b2=(ab)(a+b)(abi)(a+bi)=a2(bi)2=a2+b2a^2-b^2=(a-b)(a+b) \Rightarrow (a-bi)(a+bi)=a^2-(bi)^2=a^2+b^2).
This makes the denominator real and is similar to rationalizing the denominator with surds.
(edited 7 years ago)
Original post by timebent
for something like

35i2i\dfrac{3-5i}{2i}

why not multiply by 2i/2i why multiply by -2i/-2i instead?


Because -2i is the conjugate. Remember the conjugate of a+bi is a-bi
Reply 3
Original post by timebent
for something like

35i2i\dfrac{3-5i}{2i}

why not multiply by 2i/2i why multiply by -2i/-2i instead?


You can if you want to, infact, I'd skip the 2's and multiply by ii\frac{i}{i} myself. The importance of keeping the bi-bi in other examples is as KTA's answer.
Reply 4
Original post by Kvothe the arcane
While the conjugate of 2i is -2i, it is not necessary. As you've suggested you can multiply by 2i2i=1\dfrac{2i}{2i}=1

But where the complex number in the denominator has both a real an imaginary part, you multiply by the conjugate which is abia-bi for a complex number a+bia+bi) as you make use of the difference of two squares (a2b2=(ab)(a+b)(abi)(a+bi)=a2(bi)2=a2+b2a^2-b^2=(a-b)(a+b) \Rightarrow (a-bi)(a+bi)=a^2-(bi)^2=a^2+b^2).
This makes the denominator real and is similar to rationalizing the denominator with surds.

ok thanks
Original post by RDKGames
Because -2i is the conjugate. Remember the conjugate of a+bi is a-bi

so just because it's a conjugate that's why i'd use it?
Original post by Zacken
You can if you want to, infact, I'd skip the 2's and multiply by ii\frac{i}{i} myself. The importance of keeping the bi-bi in other examples is as KTA's answer.


so if i multiply by just i
then
i get eventually 5232i-\dfrac{5}{2} - \dfrac{3}{2} i
so am i trying to get rid of the i on the bottom or is it something else i'm trying to get rid of or change
Reply 5
Original post by timebent

so am i trying to get rid of the i on the bottom or is it something else i'm trying to get rid of or change


You just want to get rid of the isi's at the bottom so your final complex number is of the form x+iyx + iy.
Reply 6
Original post by Zacken
You just want to get rid of the isi's at the bottom so your final complex number is of the form x+iyx + iy.


ok thanks, you're so smart :biggrin:
Just like RDK, B_ with loads of number i can't remember and Kvothe :tongue:
[QUOTE=timebent;66513416]ok thanks

so just because it's a conjugate that's why i'd use it?


Yes because conjugates get rid off the imaginary parts. But as Zacken said, you can just multiply by i/i because the denominator has no real part to it. If the denominator is in the form a+bi where a and b =\= 0 then you need to multiply the numerator and denominator by the conjugate and multiplying them only by i won't get rid off the imaginary part in the denominator. Just treat these in the same way as rationalising the denominator with surds
(edited 7 years ago)
Reply 8
Original post by RDKGames
Yes because conjugates get rid off the imaginary parts. But as Zacken said, you can just multiply by i/i because the denominator has no real part to it. If the denominator is in the form a+bi where a and b =\= 0 then you need to multiply the numerator and denominator by the conjugate and multiplying them only by i won't get rid off the imaginary part in the denominator. Just treat these in the same way as rationalising the denominator with surfs


ok thanks
Are you learning FP1 and are you in year 11, going into year 12?


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Reply 10
Original post by Chittesh14
Are you learning FP1 and are you in year 11, going into year 12?


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eyy old threads
i'm in year year 12 going to 13
Original post by timebent
eyy old threads
i'm in year year 12 going to 13


Old threads ? :s-smilie:

Oh right, so are you doing Further Maths AS?
Reply 12
Original post by Chittesh14
Old threads ? :s-smilie:

Oh right, so are you doing Further Maths AS?


1 day old

No i should've done it but i didn't take it
Original post by timebent
1 day old

No i should've done it but i didn't take it


Oh right xD. Anyway, enjoy! I'm at the same stage, so let's collaborate on questions instead of asking directly on threads (because I practically do the same thing)? :smile:


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Reply 14
Original post by Chittesh14
Oh right xD. Anyway, enjoy! I'm at the same stage, so let's collaborate on questions instead of asking directly on threads (because I practically do the same thing)? :smile:


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right i'm learning from new, so currently i've gone through(supposedly) the first chapter of complex numbers, i made another thread on it xD
Original post by timebent
right i'm learning from new, so currently i've gone through(supposedly) the first chapter of complex numbers, i made another thread on it xD


Don't worry, I've finished chapter 3. We are both learning from new. I've forgot a bit of chapter 1 so I will go back. I've forgot chapter 2 I think lol and chapter 3 was just ****ing hard (a bit)


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Reply 16
Original post by Chittesh14
Don't worry, I've finished chapter 3. We are both learning from new. I've forgot a bit of chapter 1 so I will go back. I've forgot chapter 2 I think lol and chapter 3 was just ****ing hard (a bit)


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Ah check this thread, i need help on...
http://www.thestudentroom.co.uk/showthread.php?t=4222072

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